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sin×(7/15)×x>=sqrt(3)/2

  • How to use it?

  • Inequation:
  • log2(1-x)>log2(2x-8)
  • -4x^2+20x>25
  • x(x-1/2)(8+x)<0
  • (x+-3)/x+1(2x+4)_>0

  • Identical expressions

  • sin×(seven / fifteen)×x>=sqrt(three)/ two
  • sinus of ×(7 divide by 15)×x greater than or equal to square root of (3) divide by 2
  • sinus of ×(seven divide by fifteen)×x greater than or equal to square root of (three) divide by two
  • sin×(7/15)×x>=√(3)/2
  • sin×7/15×x>=sqrt3/2
  • sin×(7 divide by 15)×x>=sqrt(3) divide by 2
Solving inequalities/ sin×(7/15)×x>=sqrt(3)/2

sin×(7/15)×x>=sqrt(3)/2 inequation

A inequation with variable

The solution

You have entered [src] 
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sin(7/15)*x >= -----
                 2
$$x \sin{\left(\frac{7}{15} \right)} \geq \frac{\sqrt{3}}{2}$$ 
x*sin(7/15) >= sqrt(3)/2
Detail solution
Given the inequality:
$$x \sin{\left(\frac{7}{15} \right)} \geq \frac{\sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$x \sin{\left(\frac{7}{15} \right)} = \frac{\sqrt{3}}{2}$$
Solve:
Given the linear equation:
sin(7/15)*x = sqrt(3)/2

Expand brackets in the left part
sin7/15x = sqrt(3)/2

Expand brackets in the right part
sin7/15x = sqrt3/2

Divide both parts of the equation by sin(7/15)
x = sqrt(3)/2 / (sin(7/15))

$$x_{1} = \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
$$x_{1} = \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
This roots
$$x_{1} = \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
=
$$- \frac{1}{10} + \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
substitute to the expression
$$x \sin{\left(\frac{7}{15} \right)} \geq \frac{\sqrt{3}}{2}$$
$$\left(- \frac{1}{10} + \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}\right) \sin{\left(\frac{7}{15} \right)} \geq \frac{\sqrt{3}}{2}$$
/            ___   \                ___
|  1       \/ 3    |              \/ 3 
|- -- + -----------|*sin(7/15) >= -----
\  10   2*sin(7/15)/                2

but
/            ___   \               ___
|  1       \/ 3    |             \/ 3 
|- -- + -----------|*sin(7/15) < -----
\  10   2*sin(7/15)/               2

Then
$$x \leq \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}$$
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       x_1
Solving inequality on a graph
Rapid solution [src] 
   /     ___                \
   |   \/ 3                 |
And|----------- <= x, x < oo|
   \2*sin(7/15)             /
$$\frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}} \leq x \wedge x < \infty$$ 
(x < oo)∧(sqrt(3)/(2*sin(7/15)) <= x)
Rapid solution 2 [src] 
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    \/ 3         
[-----------, oo)
 2*sin(7/15)
$$x\ in\ \left[\frac{\sqrt{3}}{2 \sin{\left(\frac{7}{15} \right)}}, \infty\right)$$ 
x in Interval(sqrt(3)/(2*sin(7/15)), oo)
The graph
sin×(7/15)×x>=sqrt(3)/2 inequation
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